A sample of 60.0 g of formalin, containing 50.0% HCHO by mass, is treated with excess ammonia. Hexamethylenetetramine (CH2)6N4 is obtained in 85% yield based on formaldehyde. Calculate its mass in grams to two decimal places. (H 1, C 12, N 14, O 16.)

JEE Advanced levelNumerical answerDifficulty 3 of 3co004

A sample of 60.0 g of formalin, containing 50.0% $HCHO$ by mass, is treated with excess ammonia. Hexamethylenetetramine $(CH X_{2}) X_{6} N X_{4}$ is obtained in 85% yield based on formaldehyde. Calculate its mass in grams to two decimal places. (H 1, C 12, N 14, O 16.)

Numerical answer type, answer in g, tolerance 0.05.

Mass of

HCHO

= 30.0

g, i.e. 1.00 mol (M = 30). The reaction is

6 HCHO + 4 NH X_{3} (CH X_{2}) X_{6} N X_{4} + 6 H X_{2} O

, so the theoretical yield is 1.00/6 mol. The molar mass of hexamine

C X_{6} H X_{12} N X_{4}

72 + 12 + 56 = 140

g/mol. Mass

= (1.00/6) \times 140 \times 0.85 = 19.83